4MA0 Higher January 2018 Paper 3R Q18
18 There are four white tiles and three grey tiles in a bag.
Each tile has a number on one side.
There are no other tiles in the bag.
Flora takes at random a tile from the bag.
She keeps the tile.
Esther then takes at random a tile from the bag.
(a) Work out the probability that Flora takes a white tile and Esther takes a grey tile. (2)
The numbers on the tiles taken from the bag are added together to get a total \(T\).
(b) Work out the probability that \(T\) is 4 (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{4}{7} \times \dfrac{3}{6}\) | M1 |
| \(\dfrac{12}{42}\) | A1 |
| (2) |
Notes
A1: oe
| Scheme | Marks |
|---|---|
P(1, 3) = \(\dfrac{1}{7} \times \dfrac{1}{6} \times 4\left(= \dfrac{4}{42}\right)\) OR P(3, 1) = \(\dfrac{1}{7} \times \dfrac{1}{6} \times 4\left(= \dfrac{4}{42}\right)\) OR P(2, 2) = \(\dfrac{3}{7} \times \dfrac{2}{6} = \left(\dfrac{6}{42}\right)\) | M1 |
Two of P(1, 3) = \(\dfrac{1}{7} \times \dfrac{1}{6} \times 4\left(= \dfrac{4}{42}\right)\) P(3, 1) = \(\dfrac{1}{7} \times \dfrac{1}{6} \times 4\left(= \dfrac{4}{42}\right)\) P(2, 2) = \(\dfrac{3}{7} \times \dfrac{2}{6} = \left(\dfrac{6}{42}\right)\) | M1 |
| eg \(\dfrac{1}{7} \times \dfrac{1}{6} \times 4 + \dfrac{1}{7} \times \dfrac{1}{6} \times 4 + \dfrac{3}{7} \times \dfrac{2}{6}\) | M1 |
| \(\dfrac{14}{42}\) | A1 |
| (4) |
Notes
M1: for method to find P(1, 3) OR P(3, 1) OR P(2, 2)
M1: for method to find two of P(1, 3), P(3, 1), P(2, 2)
M1: for a complete method
A1: oe
SC With replacement
B2 for an answer of \(\dfrac{14}{49}\)
(B1 for \(\dfrac{1}{7} \times \dfrac{1}{7} \times 8\) or \(\dfrac{3}{7} \times \dfrac{2}{7}\))